# Why is the first return map measure preserving?

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# Definitions and Context

Let $$(X, \mc X, \mu, T)$$ be an invertible measure preserving system and let $$A$$ be a measurable subset of $$X$$ with $$\mu(A)>0$$. By Poincare recurrence we know that for almost all points $$x\in A$$ there is a positive integer $$k$$ such that $$T^kx\in A$$. Thus we may define a map $$r_A:A\to \mathbb N$$ by writing $$r_A(x) = \min\set{k\geq 1:\ T^kx\in A}$$ So $$r_A$$ is well-defined almost everywhere in $$A$$. Also define a map $$T_A:A\to A$$ by writing $$T_Ax= T^{r_A(x)}x$$ Let $$\mu_A$$ be a measure on $$A$$ obtained by restricting $$\mu$$ to $$A$$ and normalizing, that is, $$\mu_A(B)=\mu(B)/\mu(A)$$ for any measurable subset $$B$$ of $$A$$.

# The Problem

Lemma 2.43 in Einsiedler and Ward's Ergodic Theory with a view towards Number Theory states that $$T_A$$ preserves the measure $$\mu_A$$. What the proof given in the book shows is that for a measurable subset $$B$$ of $$A$$ we have $$\mu_A(T_A(B))=\mu_A(B)$$. I do not see how this shows that $$T_A$$ is measure preserving. What one needs to show is that for any measurable subset $$B$$ of $$A$$ we have $$\mu_A(T_A^{-1}(B))=\mu_A(B)$$, which is equivalent to showing that $$\mu(T_A^{-1}(B))=\mu(B)$$.

Here is what I tried. Define, as in the book, $$A_n=r_A^{-1}(n)$$. Thus the $$A_i$$'s partition $$A$$. Let $$B$$ be an arbitrary measurable subset of $$A$$. Then we have $$T_A^{-1}(B) = \bigsqcup_{n\geq 1} \set{x\in A_n:\ T_Ax\in B} = \bigsqcup_{n\geq 1} \set{x\in A_n:\ T^nx\in B}$$ Thus $$T_A^{-1}(B) = \bigsqcup_{n\geq 1} A_n\cap T^{-n}(B)$$ So we have $$\mu(T_A^{-1}(B)) = \sum_{n\geq 1} \mu(A_n\cap T^{-n}(B))$$ I am stuck here.

This fact is general for any invertible measure preserving system:

Let $$(X,\mathcal{B},\mu,T)$$ be such that $$T$$ is invertible and measurable with $$T^{-1}$$ also measurable. Then $$\mu(T^{-1}(A))=A$$ for all $$A\in\mathcal{B}$$ if and only if $$\mu(TA) = \mu(A)$$ for all $$A\in\mathcal{B}$$

In fact we only need one direction of this theorem, that $$\mu(TA)=\mu(A)$$ for all $$A\in\mathcal{B}$$ implies that $$\mu(T^{-1}(A))=A$$ for all $$A\in\mathcal{B}$$. This direction alone holds even if $$T$$ is not invertible, but is surjective. We will prove that in a minute, but this is what we need for your question.

Back to your question: Let $$B$$ be a measurable subset of $$A$$. Then since $$T$$ and $$r_A$$ are measurable, we have that $$C:=T_A^{-1}(B)$$ is also a measurable subset of $$A$$.

The proof given in the book shows that $$\mu_A(T_A C) = \mu_A(C)$$. Since $$T_A$$ is surjective (see proof below) we have $$T_A(C) = T_A T_A^{-1}(B)=B$$ which implies that $$\mu_A(B)=\mu_A(T_A^{-1}(B))$$ as required.

The proof that $$T_A$$ is surjective goes as follows: Consider the measure-preserving system $$(X,\mathcal{B},\mu,T^{-1})$$ and the same measurable set $$A$$. By poincare recurrence theorem for almost all $$a\in A$$ there exists $$n$$ such that $$T^{-n}(a)\in A$$. Let $$b:=T^{-n}(a)$$ for the minimal $$n$$, then $$T_A(b)=a$$. This shows that $$T_A$$ is essentially surjective which is enough for our purposes.

• That's a very nice way to do it. Thanks. I have just one question. The general result that you mention about $\mu(TA)=\mu(A)$ implying $T$ is measure preserving given that $T$ is surjective, isn't one assuming that the image of any measurable set under $T$ is also measurable? (I see that in our case this is not a problem.) – caffeinemachine Apr 21 '19 at 11:47
• @caffeinemachine Yes, I guess you have to assume that otherwise $\mu(TA)$ makes no sense. – Yanko Apr 21 '19 at 11:53